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robotics Designing the Mechanisms for Automated Machinery Part 14

Tham khảo tài liệu 'robotics designing the mechanisms for automated machinery part 14', kỹ thuật - công nghệ, cơ khí - chế tạo máy phục vụ nhu cầu học tập, nghiên cứu và làm việc hiệu quả | 412 Solutions to the Exercises From these formulas it follows that when the voltage is doubled us 2 we have the following expressions for the response time is 2 I tSUs Vms s s ts Ws msôsSs us ỳ Ws.ms0sSs 2 ts 0.5. From these formulas it follows that when the mass of the armature is doubled ms 2 we have the following expressions for the response time - t2 Í1 tsus I yỊ yỊĩtĩịÔịSị I us 5 Vs 2ỏsSs I us ts 1.41. 15 Solution to Exercise 4E-1 Case a From geometrical considerations the motion function n x becomes y Differentiating a we obtain X2 a Thus -X ỳ yL2-X2 b irw - I x _ Jl2-x2 c Substituting the given data into c we obtain for ỷ ỷ ----0-05.0 1 -0.0353 m sec. Jo.152-O.O52 By differentiating b we obtain the following dependence from Expression 4.3 the case where X 0 r 2 o ỷ n x x2 X2 2-x2 3 2 ------- 152 . 0-12 -0.07955m I sec2. 0.152-0.052 3 2 TEAM LRN Solutions to the Exercises 413 From c and the Relationship 4.4 we obtain Fjnput I Output n x 0.353. d Substituting the numerical data into c and d we obtain _ -5-0.05 _ Output _ -14.16 N. 0.152- 0.052 Case b From the geometry of the given mechanism we have AD - CE. Then the motion function n x is defined as follows y n 0 AO sinộ 0.2sin30 0. Im. Thus ỳ n 0 0 0.2cos30 5 0.866 m sec and y n 0 02 -0.2sin30 52 -2.5 m sec2. 16 Solution to Exercise 4E-2 From the Formula 4.24 and its derivatives we have __n 0 tan a _ . a ro n 0 Here it follows from the description of the problem that y n ự - l-cos 4 and therefore n 0 2h sin 4-0 . Thus from a we obtain 2hsin 4 0 tana -------. -----------. r hr 1 b 0.08 1-cos 4 TEAM LRN 414 Solutions to the Exercises To find the angle 0 corresponding to the maximum pressure angle amax we differentiate b or . 8ÂỈCOS40 d tana _ Y 0.08 l-COS40 -4h2 sin2 40 dộ 0.08 l- cos40 2 0. From c it follows that 2 0.08 cos 4-0 hcos 4-0 -h cos2 4-0 sin2 4-0 0 0.16COS 4-0 1-cos 4-0 On the other hand from b we have 2h sin 4-ộì-tga O.O8 l-cos4-0 0. Substituting tana tan 20 .